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elementary-differential-equations-and-boundary-value-problems-solutions-manual 3/15 Downloaded from wigs.wharton.upenn.edu on July 30, 2022 by guest engineers, economists, or chemists who need to master the prerequisites for a graduate course in mathematics. Differential Equations and Boundary Value This paper presents a simple, yet powerful approach to introducing boundary-value problems arising in electrostatics. The principles of electrostatics find numerous applications such as Linear equations (Initial Value Problems (IVP's) have unique solutions which extend to maximal intervals of existence). Consider y00+ y = 0, y(0) = 0, y(ˇ) = 1. The general solution is y= c1cosx+c2sinxand it exists on R. Also, from the boundary conditions, we have 0 = y(0) = c1but y(ˇ) = 0 6= 1. So, the BVP has no solution. Exercise 1. Boundary Value Problems is a text material on partial differential equations that teaches solutions of boundary value problems. The book also aims to build up intuition about how the solution of a problem should behave. The text consists of seven chapters. Chapter 1 covers the important topics of Fourier Series and. DOWNLOAD. TWO-DIMENSIONAL BOUNDARY VALUE PROBLEMS SIX-NODE TRIANGULAR ELEMENTS (T6) A quadratically interpolated triangular element is defined by six nodes, three at the vertices and three at the middle at each side. The middle node, depending on location, may define a straight line or a quadratic line. T6 TWO-DIMENSIONAL BOUNDARY VALUE PROBLEMS Chapter 11 Boundary Value Problems and Fourier Expansions 580 11.1 Eigenvalue Problems for y00 + λy= 0 580 11.2 Fourier Series I 586 11.3 Fourier Series II 603 Chapter 12 Fourier Solutions of Partial Differential Equations 12.1 The Heat Equation 618 12.2 The Wave Equation 630 12.3 Laplace's Equationin Rectangular Coordinates 649 If we use the conditions y(0) y ( 0) and y(2π) y ( 2 π) the only way we'll ever get a solution to the boundary value problem is if we have, y(0) = a y(2π) = a y ( 0) = a y ( 2 π) = a. for any value of a a. Also, note that if we do have these boundary conditions we'll in fact get infinitely many solutions. 3 Boundary Value Problems I Side conditions prescribing solution or derivative values at speci ed points are required to make solution of ODE unique I For initial value problem, all side conditions are speci ed at single point, say t 0 I For boundary value problem (BVP), side conditions are speci ed at more than one point I kth order ODE, or equivalent rst-order system, requires k side A boundary value problem is correctly set if it has one and only one solution within a given class of functions. Physical interpretations often suggest boundary conditions under which a problem may be correctly set. BOUNDARY VALUE PROBLEMS The basic theory of boundary value problems for ODE is more subtle than for initial value problems, and we can give only a few highlights of it here. For nota-tionalsimplicity, abbreviateboundary value problem by BVP. We begin with the two-point BVP y = f(x,y,y), a<x<b A y(a) y (a) + B y(b) y (b) = γ1 γ2 with Aand B Used to solve boundary value problems We'll look at an example 1 2 2 y dx dy ) 0 2 ( (0)1 S y y Two Steps Divide interval into steps Write differential equation in terms of values at these discrete points Solution is Desired from x=0 to /2 X=0 X= /2 Divide Interval into Pieces X0=0 X1 X2 X3 X4= /2 h Boundary Values X0=0 X1 X2 X3 X4= /2 • In a boundary-value problem, we have conditions set at two different locations • A second-order ODE d2y/dx2= g(x, y, y'), needs two boundary conditions (BC) - Simplest are y(0) = a and y(L) = b - Mixed BC: ady/dx+by = c at x = 0, L 5 Boundary-value Problems II • Solving boundary-value problems • In a boundary-value problem, we have conditions set at two different locations • A second-order ODE d2y/dx2= g(x, y, y'), needs two boundary conditions (BC) - Simplest are y(0) = a and y(L) = b - Mixed BC: ady/dx+by = c at x = 0, L 5 Boundary-value Problems II • Solving boundary-value problems Now we deal with a mixed boundary value problem for the Laplace equa- tion.: 4 (IBVP) uxx+uyy= 0 Laplace EQ. in (0,l)×(0,m) uy(x,0) = 0 NC at y= 0, u(x,m) = 0 DC at y= m, u(0,y) = g(y) nonhomogeneous DC at x= 0 u(l,y) = 0 DC at x= l Step 1. u(x,y) = X(x)Y(y) uxx= X′′Y and uyy= XY′′ Step 2.

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